Finding Basis of Null Space and Range

  • Thread starter Gooolati
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  • #1
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Homework Statement


Prove T is a linear transformation and find bases for both N(T) and R(T).


Homework Equations





The Attempt at a Solution




T:M2x3(F) [itex]\rightarrow[/itex] M2x2(F) defined by:

T(a11 a12 a13)
(a21 a22 a23)

(this is one matrix)
=

(2a11-a12 a13+2a12)
( 0 0)

(this is one matrix)

So I verified that it is a linear transformation by checking that T(cx+y)=cT(x)+T(y). But I don't understand how to find a basis for the null space and range.

I can see that since N(T)={x:T(x)=0} that N(T) here it all vectors of the form:

(t/2 t -2t)
( b b b)

(this is one matrix)

Since the 2nd row in our domain always goes to 0, the second row is arbitrary, which I represented by b.

How do I find a basis for all multiples of the matrix

t(1/2 1 -2)
( b b b)?

And I'm not even sure on how to start off finding the basis for the range. All help is appreciated. Thanks!
 

Answers and Replies

  • #2
Deveno
Science Advisor
908
6
do you have any idea of what the dimension of M2x3(F) is?

can you think of a basis for it?

can you express N(T) in terms of this basis?

alternately: do you know how to "vectorize" a matrix?

suppose you "vectorized" the elements of M2x3(F) and M2x2(F). could you pick bases for these spaces, and calculate a matrix for T in terms of those bases? what size would it have? could you find its rank and nullity?
 
  • #3
22
0
do you have any idea of what the dimension of M2x3(F) is?

Since it is 2x3, the dimension would be 6?

Because we would need 6 vectors:

\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \end{array},

\begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \end{array},

\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \end{array}

\begin{array}{ccc}
0 & 0 & 0 \\
0 & 1 & 0 \end{array},

\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \end{array}

\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \end{array}

N(T) in terms of this basis would be multiples of:

(1/2)(first matrix)+(third matrix)+(-2)(fifth matrix)?

Something doesn't seem right to be hmm.

Also, what do you mean by vectorizing?
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,228
1,830
do you have any idea of what the dimension of M2x3(F) is?

Since it is 2x3, the dimension would be 6?

Because we would need 6 vectors:

<snip>

N(T) in terms of this basis would be multiples of:

(1/2)(first matrix)+(third matrix)+(-2)(fifth matrix)?

Something doesn't seem right to be hmm.
Good start. But what about the second matrix? Doesn't T map that to 0 as well? What about the other basis elements?

Also, what do you mean by vectorizing?
You can express any matrix in M2x3 as a linear combination of those six matrices, so if you label those six matrices e1, … , e6, you could write
$$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} = a_{11}\mathbf{e}_1 + a_{21}\mathbf{e}_2 + a_{12}\mathbf{e}_3 + a_{22}\mathbf{e}_4 + a_{13}\mathbf{e}_5 + a_{23}\mathbf{e}_6$$ This matrix would therefore correspond to the coordinate vector ##(a_{11}, a_{21}, a_{12}, a_{22}, a_{13}, a_{23})^T##.
 

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